Question

A boxcar contains six complex electronic systems. Two of the six are to be randomly selected for thorough testing and then classified as defective or not defective. (Enter your probabilities as fractions.)(a)If three of the six systems are actually defective, find the probability that at least one of the two systems tested will be defective. Find the probability that both are defective. (b)If four of the six systems are actually defective, find the probability that at least one of the two systems tested will be defective. Find the probability that both are defective.

Answer 1

Answer with explanation:

Binomial probability formula to find the probability of getting success in x trials is given by :-

$P(x)=^nC_xp^x(1-p)^{n-x}$, where n is the sample size , p is the proportion of getting success in each trial.                              (1)

Given : A boxcar contains six complex electronic systems. Two of the six are to be randomly selected for thorough testing and then classified as defective or not defective.

i.e. n= 2

a) If three of the six systems are actually defective.

$\Rightarrow\ p=\dfrac{3}{6}=\dfrac{1}{2}$

Then, the probability that at least one of the two systems tested will be defective will be :_

$P(x\geq1)=1-P(x=0)\\\\=1-^{2}C_{0}(\dfrac{1}{2})^0(\dfrac{1}{2})^{2} \ \ \text{[By using formula (1)]}\\\\=1-(\dfrac{1}{2})^2\ \ \because\ ^nC_0=1\\\\=1-\dfrac{1}{4}=\dfrac{3}{4}$

Hence, the probability that at least one of the two systems tested will be defective  = $\dfrac{3}{4}$

b) If four of the six systems are actually defective.

$\Rightarrow\ p=\dfrac{4}{6}=\dfrac{2}{3}$

Then, the probability that at least one of the two systems tested will be defective will be :_

$P(x\geq1)=1-P(x=0)\\\\=1-^{2}C_{0}(\dfrac{2}{3})^0(1-\dfrac{2}{3})^{2} \ \ \text{[By using formula (1)]}\\\\=1-(\dfrac{1}{3})^2\ \ \because\ ^nC_0=1\\\\=1-\dfrac{1}{9}=\dfrac{8}{9}$

Hence, the probability that at least one of the two systems tested will be defective  =$\dfrac{8}{9}$

The probability that both are defective :

$P(x=2)=^{2}C_{2}(\dfrac{2}{3})^2(1-\dfrac{2}{3})^{0} \ \ \text{[By using formula (1)]}\\\\=(\dfrac{2}{3})^2\ \ \because\ ^nC_n=1\\\\=\dfrac{4}{9}$

Hence, the probability that both are defective = $\dfrac{4}{9}$